Exercise 6.3.4

$|\mathit{sin}(x)|\le 1$ and so $\left|h_n(x)\right|<\frac{1}{\sqrt{n}}$ which shows that $h_n\to 0$ uniformly on $R$. $h_n^{\prime }(x)=\frac{n\cos <!--\; FUNCTION\; APPLICATION\; -->(\mathit{nx})}{\sqrt{n}}=\sqrt{n}\cos <!--\; FUNCTION\; APPLICATION\; -->(\mathit{nx})$. Intuitively this diverges because of the unbounded $\sqrt{n}$ factor, but to prove it formally requires some thought. We want to show that for any fixed real numbers $x$ and $M$ we can find some $n$ where $h_n^{\prime }(x)\ge M$ (this will show that $h_n^{\prime }(x)$ is unbounded and thus diverges). First let $N_1>4M^2$, then we just need to find some $n\ge N_1$ so that $\left|\mathit{nx}-\mathit{Z\pi }\right|\le \pi ∕3$ for some integer $Z$; this would cause $\left|\cos <!--\; FUNCTION\; APPLICATION\; -->(\mathit{nx})\right|\ge 1∕2$ and thus $h_n^{\prime }(x)\ge M$.

Express $N_{1}x=2\mathit{\pi P}+k$ where $P$ is some integer and $0\le k<2\pi$. (This next bit is reminiscent of arithmetic modulo $2\pi$.) Now if $k\in [0,\pi ∕3]$ or $[2\pi ∕3,4\pi ∕3]$ or $[5\pi ∕3,2\pi ]$ we’re done. Otherwise $k\in (\pi ∕3,2\pi ∕3)$ or $(4\pi ∕3,5\pi ∕3)$; so consider $2N_{1}x=4\mathit{\pi P}+2k$, with $2k\in (2\pi ∕3,4\pi ∕3)$ or $(8\pi ∕3,10\pi ∕3)$; both of these cases will have $2N_{1}x$ within $\pi ∕3$ of a multiple of $\pi$, hence $h_n^{\prime }(x)$ diverges for all $x$.